Continuous time to Discrete time using periodic sampling
From Schaum's outlines, Digital Signal Processing, 2nd Edition, 2012, page 114:
Consider the discrete time-sequence:
$$x[n] = cos(frac{pi}{8}n)$$
Find two different continuous time signals:
$$X_a(t) = cos(2 pi f_0 t)$$
that would produce this sequence when sampled at a frequency of $$f_s = 10 KHz$$
So I start out with Ideal A/D converter relationship for converting from continuous to discrete time:
$$x[n]=X_a(t=nT_s)$$
$$x[n]= cos(2 pi f_0 n T_s)$$
$$T_s = 1 / f_s $$
$$x[n]= cos(2 pi n frac{f_0}{f_s})$$
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k)$$
$$x[n]= cos(frac{2 pi n f_0}{f_s} + frac{2 pi k f_s}{f_s})$$
$$x[n]= cos(frac{2 pi n f_0 + 2 pi k f_s}{f_s})$$
$$x[n]= cos(2 pi frac{n f_0 + k f_s}{f_s})$$
Here's where I have the problem. The book says at this point I should have this instead:
$$x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$$
and:
$$f = f_0 + k f_s $$
How did the book get that result instead?
fourier-transform
asked Nov 23 '18 at 14:41
Bill MooreBill Moore
1176
add a comment |
From Schaum's outlines, Digital Signal Processing, 2nd Edition, 2012, page 114:
Consider the discrete time-sequence:
$$x[n] = cos(frac{pi}{8}n)$$
Find two different continuous time signals:
$$X_a(t) = cos(2 pi f_0 t)$$
that would produce this sequence when sampled at a frequency of $$f_s = 10 KHz$$
So I start out with Ideal A/D converter relationship for converting from continuous to discrete time:
$$x[n]=X_a(t=nT_s)$$
$$x[n]= cos(2 pi f_0 n T_s)$$
$$T_s = 1 / f_s $$
$$x[n]= cos(2 pi n frac{f_0}{f_s})$$
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k)$$
$$x[n]= cos(frac{2 pi n f_0}{f_s} + frac{2 pi k f_s}{f_s})$$
$$x[n]= cos(frac{2 pi n f_0 + 2 pi k f_s}{f_s})$$
$$x[n]= cos(2 pi frac{n f_0 + k f_s}{f_s})$$
Here's where I have the problem. The book says at this point I should have this instead:
$$x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$$
and:
$$f = f_0 + k f_s $$
How did the book get that result instead?
fourier-transform
asked Nov 23 '18 at 14:41
Bill MooreBill Moore
1176
You added a phase offset (it doesn't vary with $n$), and the book added a frequency offset (it does vary with $n$). Both are legitimate answers to the question you presented, but the frequency offset is really meant to get you thinking about aliases. The DFT lives on the unit circle of the z-plane. Every $2pi$ radians trip around the unit circle is $f_s$ Hz in frequency, and is an alias for where you started.
– Andy Walls
Nov 24 '18 at 4:11
add a comment |
From Schaum's outlines, Digital Signal Processing, 2nd Edition, 2012, page 114:
Consider the discrete time-sequence:
$$x[n] = cos(frac{pi}{8}n)$$
Find two different continuous time signals:
$$X_a(t) = cos(2 pi f_0 t)$$
that would produce this sequence when sampled at a frequency of $$f_s = 10 KHz$$
So I start out with Ideal A/D converter relationship for converting from continuous to discrete time:
$$x[n]=X_a(t=nT_s)$$
$$x[n]= cos(2 pi f_0 n T_s)$$
$$T_s = 1 / f_s $$
$$x[n]= cos(2 pi n frac{f_0}{f_s})$$
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k)$$
$$x[n]= cos(frac{2 pi n f_0}{f_s} + frac{2 pi k f_s}{f_s})$$
$$x[n]= cos(frac{2 pi n f_0 + 2 pi k f_s}{f_s})$$
$$x[n]= cos(2 pi frac{n f_0 + k f_s}{f_s})$$
Here's where I have the problem. The book says at this point I should have this instead:
$$x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$$
and:
$$f = f_0 + k f_s $$
How did the book get that result instead?
fourier-transform
asked Nov 23 '18 at 14:41
Bill MooreBill Moore
1176
From Schaum's outlines, Digital Signal Processing, 2nd Edition, 2012, page 114:
Consider the discrete time-sequence:
$$x[n] = cos(frac{pi}{8}n)$$
Find two different continuous time signals:
$$X_a(t) = cos(2 pi f_0 t)$$
that would produce this sequence when sampled at a frequency of $$f_s = 10 KHz$$
So I start out with Ideal A/D converter relationship for converting from continuous to discrete time:
$$x[n]=X_a(t=nT_s)$$
$$x[n]= cos(2 pi f_0 n T_s)$$
$$T_s = 1 / f_s $$
$$x[n]= cos(2 pi n frac{f_0}{f_s})$$
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k)$$
$$x[n]= cos(frac{2 pi n f_0}{f_s} + frac{2 pi k f_s}{f_s})$$
$$x[n]= cos(frac{2 pi n f_0 + 2 pi k f_s}{f_s})$$
$$x[n]= cos(2 pi frac{n f_0 + k f_s}{f_s})$$
Here's where I have the problem. The book says at this point I should have this instead:
$$x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$$
and:
$$f = f_0 + k f_s $$
How did the book get that result instead?
fourier-transform
fourier-transform
asked Nov 23 '18 at 14:41
Bill MooreBill Moore
1176
asked Nov 23 '18 at 14:41
Bill MooreBill Moore
1176
edited Nov 23 '18 at 17:02
Bill Moore
asked Nov 23 '18 at 14:41
Bill MooreBill Moore
1176
asked Nov 23 '18 at 14:41
Bill MooreBill Moore
1176
asked Nov 23 '18 at 14:41
Bill MooreBill Moore
1176
1176
You added a phase offset (it doesn't vary with $n$), and the book added a frequency offset (it does vary with $n$). Both are legitimate answers to the question you presented, but the frequency offset is really meant to get you thinking about aliases. The DFT lives on the unit circle of the z-plane. Every $2pi$ radians trip around the unit circle is $f_s$ Hz in frequency, and is an alias for where you started.
– Andy Walls
Nov 24 '18 at 4:11
add a comment |
You added a phase offset (it doesn't vary with $n$), and the book added a frequency offset (it does vary with $n$). Both are legitimate answers to the question you presented, but the frequency offset is really meant to get you thinking about aliases. The DFT lives on the unit circle of the z-plane. Every $2pi$ radians trip around the unit circle is $f_s$ Hz in frequency, and is an alias for where you started.
– Andy Walls
Nov 24 '18 at 4:11
You added a phase offset (it doesn't vary with $n$), and the book added a frequency offset (it does vary with $n$). Both are legitimate answers to the question you presented, but the frequency offset is really meant to get you thinking about aliases. The DFT lives on the unit circle of the z-plane. Every $2pi$ radians trip around the unit circle is $f_s$ Hz in frequency, and is an alias for where you started.
– Andy Walls
Nov 24 '18 at 4:11
You added a phase offset (it doesn't vary with $n$), and the book added a frequency offset (it does vary with $n$). Both are legitimate answers to the question you presented, but the frequency offset is really meant to get you thinking about aliases. The DFT lives on the unit circle of the z-plane. Every $2pi$ radians trip around the unit circle is $f_s$ Hz in frequency, and is an alias for where you started.
– Andy Walls
Nov 24 '18 at 4:11
add a comment |
2 Answers
2
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The problem lies where you added $2pi k$. Instead it should be $2pi k n$. So that becomes $x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi kn)$ which gives $x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$.
answered Nov 23 '18 at 14:53
Himanshu SharmaHimanshu Sharma
113
that's a good point! thanks.
– Bill Moore
Nov 23 '18 at 14:56
wait, can you do that? Wouldn't that mean the additive offset of the (2 pi k) is changing with n if you multiply it by n? consider: cos(x) = cos(x + 2 pi)
– Bill Moore
Nov 23 '18 at 15:01
In discrete signals, frequency range is from $(-pi, pi)$. What they are trying to say is that a frequency of $f_{o}+kf_{s}$ will produce the same aliasing effect as does the frequence $f_{o}$ alone in itself. For example, take $x(n)=cos(0.4pi n)=cos(2pifrac{f_{o}}{f_{s}}n) implies frac{f_{o}}{f_{s}} = 0.2$. Let me add $k=1$ to $frac{f_{o}}{f_{s}}$. The cosine becomes $cos(1.2times 2pi n) = cos(2.4pi n) = cos(2pi n - 0.4pi n) = cos(0.4pi n)$. I hope you understand this, because its very important. You should try with other values of $k$ also. Feel free to question.
– Himanshu Sharma
Nov 23 '18 at 15:48
The book is not teaching you periodicity, mind that. The section of book that you are reading is concerned about sampling of the continuous time signals to discrete time. They are teaching you what is called aliasing. Do surf about it on the internet, you will understand what the book and I am trying to tell you. Keeping periodicity in mind, you may be correct, but that is not what the text is trying to achieve there. Hope that helps.
– Himanshu Sharma
Nov 23 '18 at 15:53
its starting to make sense, because $$cos(x) = cos(x + 2 pi k n)$$ The phase offset can be any integer value multiplied by $$ 2 pi $$. It doesn't matter if you multiply 2 pi by just k or by (k*n), they are both still integers, and the resultant offset is still a period of cos
– Bill Moore
Nov 23 '18 at 15:58
|
show 1 more comment
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k n)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + k)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + frac{k}{f_s})$$
$$x[n]= cos(2 pi (n frac{1}{f_s}) (f_0 + k f_s))$$
$$x[n]= cos(2 pi t (f_0 + k f_s))$$
$$f = f_0 + k f_s $$
answered Nov 23 '18 at 16:28
Bill MooreBill Moore
1176
add a comment |
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2 Answers
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2 Answers
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The problem lies where you added $2pi k$. Instead it should be $2pi k n$. So that becomes $x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi kn)$ which gives $x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$.
answered Nov 23 '18 at 14:53
Himanshu SharmaHimanshu Sharma
113
that's a good point! thanks.
– Bill Moore
Nov 23 '18 at 14:56
wait, can you do that? Wouldn't that mean the additive offset of the (2 pi k) is changing with n if you multiply it by n? consider: cos(x) = cos(x + 2 pi)
– Bill Moore
Nov 23 '18 at 15:01
In discrete signals, frequency range is from $(-pi, pi)$. What they are trying to say is that a frequency of $f_{o}+kf_{s}$ will produce the same aliasing effect as does the frequence $f_{o}$ alone in itself. For example, take $x(n)=cos(0.4pi n)=cos(2pifrac{f_{o}}{f_{s}}n) implies frac{f_{o}}{f_{s}} = 0.2$. Let me add $k=1$ to $frac{f_{o}}{f_{s}}$. The cosine becomes $cos(1.2times 2pi n) = cos(2.4pi n) = cos(2pi n - 0.4pi n) = cos(0.4pi n)$. I hope you understand this, because its very important. You should try with other values of $k$ also. Feel free to question.
– Himanshu Sharma
Nov 23 '18 at 15:48
The book is not teaching you periodicity, mind that. The section of book that you are reading is concerned about sampling of the continuous time signals to discrete time. They are teaching you what is called aliasing. Do surf about it on the internet, you will understand what the book and I am trying to tell you. Keeping periodicity in mind, you may be correct, but that is not what the text is trying to achieve there. Hope that helps.
– Himanshu Sharma
Nov 23 '18 at 15:53
its starting to make sense, because $$cos(x) = cos(x + 2 pi k n)$$ The phase offset can be any integer value multiplied by $$ 2 pi $$. It doesn't matter if you multiply 2 pi by just k or by (k*n), they are both still integers, and the resultant offset is still a period of cos
– Bill Moore
Nov 23 '18 at 15:58
|
show 1 more comment
The problem lies where you added $2pi k$. Instead it should be $2pi k n$. So that becomes $x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi kn)$ which gives $x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$.
answered Nov 23 '18 at 14:53
Himanshu SharmaHimanshu Sharma
113
that's a good point! thanks.
– Bill Moore
Nov 23 '18 at 14:56
wait, can you do that? Wouldn't that mean the additive offset of the (2 pi k) is changing with n if you multiply it by n? consider: cos(x) = cos(x + 2 pi)
– Bill Moore
Nov 23 '18 at 15:01
In discrete signals, frequency range is from $(-pi, pi)$. What they are trying to say is that a frequency of $f_{o}+kf_{s}$ will produce the same aliasing effect as does the frequence $f_{o}$ alone in itself. For example, take $x(n)=cos(0.4pi n)=cos(2pifrac{f_{o}}{f_{s}}n) implies frac{f_{o}}{f_{s}} = 0.2$. Let me add $k=1$ to $frac{f_{o}}{f_{s}}$. The cosine becomes $cos(1.2times 2pi n) = cos(2.4pi n) = cos(2pi n - 0.4pi n) = cos(0.4pi n)$. I hope you understand this, because its very important. You should try with other values of $k$ also. Feel free to question.
– Himanshu Sharma
Nov 23 '18 at 15:48
The book is not teaching you periodicity, mind that. The section of book that you are reading is concerned about sampling of the continuous time signals to discrete time. They are teaching you what is called aliasing. Do surf about it on the internet, you will understand what the book and I am trying to tell you. Keeping periodicity in mind, you may be correct, but that is not what the text is trying to achieve there. Hope that helps.
– Himanshu Sharma
Nov 23 '18 at 15:53
its starting to make sense, because $$cos(x) = cos(x + 2 pi k n)$$ The phase offset can be any integer value multiplied by $$ 2 pi $$. It doesn't matter if you multiply 2 pi by just k or by (k*n), they are both still integers, and the resultant offset is still a period of cos
– Bill Moore
Nov 23 '18 at 15:58
|
show 1 more comment
The problem lies where you added $2pi k$. Instead it should be $2pi k n$. So that becomes $x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi kn)$ which gives $x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$.
answered Nov 23 '18 at 14:53
Himanshu SharmaHimanshu Sharma
113
The problem lies where you added $2pi k$. Instead it should be $2pi k n$. So that becomes $x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi kn)$ which gives $x[n]= cos(2 pi frac{f_0 + k f_s}{f_s} n)$.
answered Nov 23 '18 at 14:53
Himanshu SharmaHimanshu Sharma
113
answered Nov 23 '18 at 14:53
Himanshu SharmaHimanshu Sharma
113
answered Nov 23 '18 at 14:53
Himanshu SharmaHimanshu Sharma
113
answered Nov 23 '18 at 14:53
Himanshu SharmaHimanshu Sharma
113
113
that's a good point! thanks.
– Bill Moore
Nov 23 '18 at 14:56
wait, can you do that? Wouldn't that mean the additive offset of the (2 pi k) is changing with n if you multiply it by n? consider: cos(x) = cos(x + 2 pi)
– Bill Moore
Nov 23 '18 at 15:01
In discrete signals, frequency range is from $(-pi, pi)$. What they are trying to say is that a frequency of $f_{o}+kf_{s}$ will produce the same aliasing effect as does the frequence $f_{o}$ alone in itself. For example, take $x(n)=cos(0.4pi n)=cos(2pifrac{f_{o}}{f_{s}}n) implies frac{f_{o}}{f_{s}} = 0.2$. Let me add $k=1$ to $frac{f_{o}}{f_{s}}$. The cosine becomes $cos(1.2times 2pi n) = cos(2.4pi n) = cos(2pi n - 0.4pi n) = cos(0.4pi n)$. I hope you understand this, because its very important. You should try with other values of $k$ also. Feel free to question.
– Himanshu Sharma
Nov 23 '18 at 15:48
The book is not teaching you periodicity, mind that. The section of book that you are reading is concerned about sampling of the continuous time signals to discrete time. They are teaching you what is called aliasing. Do surf about it on the internet, you will understand what the book and I am trying to tell you. Keeping periodicity in mind, you may be correct, but that is not what the text is trying to achieve there. Hope that helps.
– Himanshu Sharma
Nov 23 '18 at 15:53
its starting to make sense, because $$cos(x) = cos(x + 2 pi k n)$$ The phase offset can be any integer value multiplied by $$ 2 pi $$. It doesn't matter if you multiply 2 pi by just k or by (k*n), they are both still integers, and the resultant offset is still a period of cos
– Bill Moore
Nov 23 '18 at 15:58
|
show 1 more comment
that's a good point! thanks.
– Bill Moore
Nov 23 '18 at 14:56
wait, can you do that? Wouldn't that mean the additive offset of the (2 pi k) is changing with n if you multiply it by n? consider: cos(x) = cos(x + 2 pi)
– Bill Moore
Nov 23 '18 at 15:01
In discrete signals, frequency range is from $(-pi, pi)$. What they are trying to say is that a frequency of $f_{o}+kf_{s}$ will produce the same aliasing effect as does the frequence $f_{o}$ alone in itself. For example, take $x(n)=cos(0.4pi n)=cos(2pifrac{f_{o}}{f_{s}}n) implies frac{f_{o}}{f_{s}} = 0.2$. Let me add $k=1$ to $frac{f_{o}}{f_{s}}$. The cosine becomes $cos(1.2times 2pi n) = cos(2.4pi n) = cos(2pi n - 0.4pi n) = cos(0.4pi n)$. I hope you understand this, because its very important. You should try with other values of $k$ also. Feel free to question.
– Himanshu Sharma
Nov 23 '18 at 15:48
The book is not teaching you periodicity, mind that. The section of book that you are reading is concerned about sampling of the continuous time signals to discrete time. They are teaching you what is called aliasing. Do surf about it on the internet, you will understand what the book and I am trying to tell you. Keeping periodicity in mind, you may be correct, but that is not what the text is trying to achieve there. Hope that helps.
– Himanshu Sharma
Nov 23 '18 at 15:53
its starting to make sense, because $$cos(x) = cos(x + 2 pi k n)$$ The phase offset can be any integer value multiplied by $$ 2 pi $$. It doesn't matter if you multiply 2 pi by just k or by (k*n), they are both still integers, and the resultant offset is still a period of cos
– Bill Moore
Nov 23 '18 at 15:58
that's a good point! thanks.
– Bill Moore
Nov 23 '18 at 14:56
that's a good point! thanks.
– Bill Moore
Nov 23 '18 at 14:56
wait, can you do that? Wouldn't that mean the additive offset of the (2 pi k) is changing with n if you multiply it by n? consider: cos(x) = cos(x + 2 pi)
– Bill Moore
Nov 23 '18 at 15:01
wait, can you do that? Wouldn't that mean the additive offset of the (2 pi k) is changing with n if you multiply it by n? consider: cos(x) = cos(x + 2 pi)
– Bill Moore
Nov 23 '18 at 15:01
In discrete signals, frequency range is from $(-pi, pi)$. What they are trying to say is that a frequency of $f_{o}+kf_{s}$ will produce the same aliasing effect as does the frequence $f_{o}$ alone in itself. For example, take $x(n)=cos(0.4pi n)=cos(2pifrac{f_{o}}{f_{s}}n) implies frac{f_{o}}{f_{s}} = 0.2$. Let me add $k=1$ to $frac{f_{o}}{f_{s}}$. The cosine becomes $cos(1.2times 2pi n) = cos(2.4pi n) = cos(2pi n - 0.4pi n) = cos(0.4pi n)$. I hope you understand this, because its very important. You should try with other values of $k$ also. Feel free to question.
– Himanshu Sharma
Nov 23 '18 at 15:48
In discrete signals, frequency range is from $(-pi, pi)$. What they are trying to say is that a frequency of $f_{o}+kf_{s}$ will produce the same aliasing effect as does the frequence $f_{o}$ alone in itself. For example, take $x(n)=cos(0.4pi n)=cos(2pifrac{f_{o}}{f_{s}}n) implies frac{f_{o}}{f_{s}} = 0.2$. Let me add $k=1$ to $frac{f_{o}}{f_{s}}$. The cosine becomes $cos(1.2times 2pi n) = cos(2.4pi n) = cos(2pi n - 0.4pi n) = cos(0.4pi n)$. I hope you understand this, because its very important. You should try with other values of $k$ also. Feel free to question.
– Himanshu Sharma
Nov 23 '18 at 15:48
The book is not teaching you periodicity, mind that. The section of book that you are reading is concerned about sampling of the continuous time signals to discrete time. They are teaching you what is called aliasing. Do surf about it on the internet, you will understand what the book and I am trying to tell you. Keeping periodicity in mind, you may be correct, but that is not what the text is trying to achieve there. Hope that helps.
– Himanshu Sharma
Nov 23 '18 at 15:53
The book is not teaching you periodicity, mind that. The section of book that you are reading is concerned about sampling of the continuous time signals to discrete time. They are teaching you what is called aliasing. Do surf about it on the internet, you will understand what the book and I am trying to tell you. Keeping periodicity in mind, you may be correct, but that is not what the text is trying to achieve there. Hope that helps.
– Himanshu Sharma
Nov 23 '18 at 15:53
its starting to make sense, because $$cos(x) = cos(x + 2 pi k n)$$ The phase offset can be any integer value multiplied by $$ 2 pi $$. It doesn't matter if you multiply 2 pi by just k or by (k*n), they are both still integers, and the resultant offset is still a period of cos
– Bill Moore
Nov 23 '18 at 15:58
its starting to make sense, because $$cos(x) = cos(x + 2 pi k n)$$ The phase offset can be any integer value multiplied by $$ 2 pi $$. It doesn't matter if you multiply 2 pi by just k or by (k*n), they are both still integers, and the resultant offset is still a period of cos
– Bill Moore
Nov 23 '18 at 15:58
|
show 1 more comment
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k n)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + k)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + frac{k}{f_s})$$
$$x[n]= cos(2 pi (n frac{1}{f_s}) (f_0 + k f_s))$$
$$x[n]= cos(2 pi t (f_0 + k f_s))$$
$$f = f_0 + k f_s $$
answered Nov 23 '18 at 16:28
Bill MooreBill Moore
1176
add a comment |
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k n)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + k)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + frac{k}{f_s})$$
$$x[n]= cos(2 pi (n frac{1}{f_s}) (f_0 + k f_s))$$
$$x[n]= cos(2 pi t (f_0 + k f_s))$$
$$f = f_0 + k f_s $$
answered Nov 23 '18 at 16:28
Bill MooreBill Moore
1176
add a comment |
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k n)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + k)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + frac{k}{f_s})$$
$$x[n]= cos(2 pi (n frac{1}{f_s}) (f_0 + k f_s))$$
$$x[n]= cos(2 pi t (f_0 + k f_s))$$
$$f = f_0 + k f_s $$
answered Nov 23 '18 at 16:28
Bill MooreBill Moore
1176
Since Cos is periodic we have:
$$x[n]= cos(frac{2 pi n f_0}{f_s} + 2 pi k n)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + k)$$
$$x[n]= cos(2 pi n frac{f_0}{f_s} + frac{k}{f_s})$$
$$x[n]= cos(2 pi (n frac{1}{f_s}) (f_0 + k f_s))$$
$$x[n]= cos(2 pi t (f_0 + k f_s))$$
$$f = f_0 + k f_s $$
answered Nov 23 '18 at 16:28
Bill MooreBill Moore
1176
answered Nov 23 '18 at 16:28
Bill MooreBill Moore
1176
answered Nov 23 '18 at 16:28
Bill MooreBill Moore
1176
answered Nov 23 '18 at 16:28
Bill MooreBill Moore
1176
1176
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You added a phase offset (it doesn't vary with $n$), and the book added a frequency offset (it does vary with $n$). Both are legitimate answers to the question you presented, but the frequency offset is really meant to get you thinking about aliases. The DFT lives on the unit circle of the z-plane. Every $2pi$ radians trip around the unit circle is $f_s$ Hz in frequency, and is an alias for where you started.
– Andy Walls
Nov 24 '18 at 4:11